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I want to compute following radical
$$\sqrt[4]{28+16 \sqrt 3}$$
For that, I first tried to rewrite this in terms of exponential.
$$(28+16\cdot 3^{\frac{1}{2}})^{\frac{1}{4}}$$
We know that $ 28 = 2 \cdot 7^{\frac{1}{2}}$
$$(2 \cdot 7^{\frac{1}{2}}+16\cdot 3^{\frac{1}{2}})^{\frac{1}{4}}$$
However, I'm stuck at this step. Could you assist me?
Regards
$$\sqrt[4] {28+16\sqrt 3}=\sqrt[4] {(\sqrt {12})^2 +(\sqrt {16})^2 +2\sqrt {16\cdot 12}}=\sqrt {4+2\sqrt 3}=\sqrt {(\sqrt 3)^2 +(1)^2 +2\sqrt {3\cdot 1}}=\sqrt 3 +1$$
Hint.
$28+16\sqrt3=12+16+2\cdot2\sqrt3\cdot4=(2\sqrt3)^2+4^2+2\cdot2\sqrt3\cdot4=(4+2\sqrt3)^2$
$4+2\sqrt3=3+1+2\cdot1\cdot\sqrt3=(\sqrt3)^2+1^2+2\cdot1\cdot\sqrt3=(\sqrt3+1)^2$
Hint:
Try to write $$28+16 \sqrt 3=(a+b\sqrt 3)^4$$ for suitable $a$ and $b$.
You obtain, reordering the terms \begin{align} (a+b\sqrt 3)^4&=(a^4+18a^2b^2+9b^4)+4ab(a^2+3b^2)\sqrt 3.\\ \end{align} Can you choose $a$ and $b$ so that $$a^4+18a^2b^2+9b^4=28,\quad ab(a^2+3b^2)=4?$$
To find the fourth root, we need to use Bill Dubuque's denesting algorithm for square roots twice.
First time round, we can find that $\sqrt{\text{norm}} = \sqrt{28^2 - 3 \cdot 16^2} = \sqrt{16} = 4$. Subtracting out the norm gives $24 + 16 \sqrt{3}$, and when you divide this by $\sqrt{\text{trace}} = \sqrt{2 \cdot 24} = 4 \sqrt{3}$, we get $\frac{8 \sqrt{3}\sqrt 3}{4 \sqrt 3} + \frac{16 \sqrt3}{4 \sqrt3} = 2\sqrt{3}+4 = 4 + 2\sqrt{3}$.
Second time round, we can find that $\sqrt{\text{norm}} = \sqrt{4^2 - 3 \cdot 2^2 -8} = \sqrt{4} = 2$. Subtracting out the norm gives $2 + 2\sqrt{3}$, and when you divide this by $\sqrt{\text{trace}} = \sqrt{4} = 2$, you get $1 + \sqrt{3}$.
Note
$$\sqrt[4]{28+16 \sqrt 3}=\sqrt[4]{4(7+2 \sqrt {3\cdot4})} = \sqrt[4]{4(\sqrt4+\sqrt {3})^2}\\ = \sqrt{4+2\sqrt{1\cdot3}}=\sqrt{(1+\sqrt3)^2}=1+\sqrt3 $$
In general, there's not much you can do with an expression like that. But if you know that the answer is in the form $a+b\sqrt 3$ for integers $a$ and $b$, then you can solve it like this:
Step 1
($a+b\sqrt 3)^4=28+16\sqrt 3$. So suppose ($a+b\sqrt 3)^2=c+d\sqrt 3$. Then $(c+d\sqrt 3)^2=28+16\sqrt 3$, i.e.
$$(c^2+3d^2)+2cd\sqrt 3=28+16\sqrt 3$$
$c^2+3d^2$ and $2cd$ are integers, so this can only be true if $c^2+3d^2=28$ and $2cd=16$.
Then $d=8/c$, so $c^2+\dfrac{192}{c^2}=28$. Multiplying by $c^2$ and rearranging, we get a quadratic in $c^2$:
$$c^4-28c^2+192=0$$
This factors as $(c^2-12)(c^2-16)=0$. Since $c$ is an integer, we must have $c=\pm4$, and so $d=\pm2$. So we have two possible solutions: $\pm(4+2\sqrt 3)$. (You should try squaring this to check that you get the expected result $28+16\sqrt 3$). Pick the positive solution $4+2\sqrt 3$ (if you choose the negative solution, you will find that Step 2 fails).
Step 2
Now you need to solve $(a+b\sqrt 3)^2=4+2\sqrt 3$. You should be able to do this yourself now, using exactly the same method as Step 1.
Alternatively, you can use Michael Rozenberg's identities, where this identity applies to the question:
$$\sqrt{a+\sqrt{b}}=\sqrt{\frac{a+\sqrt{a^2-b}}{2}}+\sqrt{\frac{a-\sqrt{a^2-b}}{2}}$$
We have $a = 28$, $b = 16^2 \cdot 3 = 768$. Applying this identity gives:
$$\sqrt{28+\sqrt{768}}=\sqrt{\frac{28+\sqrt{16}}{2}}+\sqrt{\frac{28-\sqrt{16}}{2}}=\sqrt{16}+\sqrt{12}=4+2\sqrt{3}.$$
Applying the identity once more gives, where $a=4, b=2^2 \cdot 3 = 12$ gives: $$\sqrt{4+\sqrt{12}}=\sqrt{\frac{4+\sqrt{4^2-12}}{2}}+\sqrt{\frac{4-\sqrt{4^2-12}}{2}}=\sqrt{\frac{4+2}{2}}+\sqrt{\frac{4-2}{2}}=\sqrt{3}+1.$$
